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In the question number 52, the speed with which the stone hits the ground is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> m `s^(-1)` <br/>90 m `s^(-1)` <br/>99 m `s^(-1)` <br/>49 m `s^(-1)` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Motion along horizontal <a href="https://interviewquestions.tuteehub.com/tag/direction-1696" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTION">DIRECTION</a>, `downarrow +ve` <br/> `u_(x) = 15 ms^(-1), a_(x) = 0`<br/> `v_(x) = u_(x) + a_(x)t = 15 + 0 xx 10 = 15 ms^(-1)`<br/> Motion along vertical direction, <br/> `u_(y) = 0, a_(y) = <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> `v_(y) = u_(y) + a_(y)t = 0 + 9.8 xx 10 = 98 ms^(-1)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Speed of the stone when it hits the ground is <br/> `v = sqrt(v_(x)^(2) + v_(y)^(2)) = sqrt((15)^(2) + (98)^(2)) ~~ 99 ms^(-1)`</body></html> | |